TITLE: Alternative ways to prove that $\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$ without complex analysis?
QUESTION [6 upvotes]: $$\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx=2\pi$$
I derived this rather incredible result via Cauchy's theorem as I was working through some simple contour integrals. I was wondering if this integral can be solved without complex analysis, and if so how? I can't see any obvious substitutions that can be possible, nor any parametrisations that could simplify the integral through Feynman's method.

REPLY [4 votes]: \begin{align}
\int_0^{2\pi} \cos(\sin{x})e^{\cos{x}} dx &=Re\left( \int_0^{2\pi} e^{i\sin(x)}e^{\cos{x}} dx\right) \\
&= Re \left(\int_0^{2\pi} e^{e^{i x}} dx\right)\\
&= Re \left(\sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} e^{inx} dx\right) \\
&= \sum_{n=0}^\infty \frac{1}{n!}\int_0^{2\pi} \cos(nx) dx \\
&= \int_0^{2\pi} dx  = 2\pi
\end{align}